Poj 2765 decimal places (Accuracy Problem)

2765: octal decimal

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Total time limit:

1000 ms

Memory limit:

65536kb

Description

Decimal decimal places can be exactly expressed. For example, 0.75 in octal equals 0.963125 in decimal format (7/8 + 5/64 ). All octal decimal places with N digits after the decimal point can be expressed as decimal places with no more than 3N digits after the decimal point.

Your task is to write a program to convert the octal decimal places (0, 1) into decimal places.

Input

The input includes several octal decimal places, each occupying a row. Each decimal point is in the form of 0. d1d2d3... DK. Here Di is the eight decimal number 0... 7, and 0 <k <15 is known.

Output

For each input octal decimal point, enter a row in the form below

0. d1d2d3... DK [8] = 0. d1d2d3... DM [10]

Here, the left side is the input decimal place, and the right side is an equal decimal place. The output decimal point cannot end with 0, that is, DM is not equal to 0.

Sample Input

0.750.00010.01234567

Sample output

0.75 [8] = 0.953125 [10]0.0001 [8] = 0.000244140625 [10]0.01234567 [8] = 0.020408093929290771484375 [10]

Prompt

If you use a string to read decimal digits, you can abort the input as follows:

Char octal [100];
While (CIN> octal ){
...
}

This question should be a basic question. When I first saw this question, I thought it was a high-precision question. I didn't dare to use double in the beginning, later, I saw that the 8-digit number range was only 15. So I thought that double should also be able to do it. I took a look at other people's ideas. It's so clever that I can directly process it in reverse order of strings, it is worth learning to convert an octal decimal number into a decimal integer ~

The following is the code. The key to being brief is to master this kind of thinking, which is a clever way of writing;

It is different from the common method in the past;

0.75 (8) = 5/64 + 7/8;

# Include <cstdio> # include <cstring> int main () {char s [20]; int Len; Double N; while (scanf ("% s", S )! = EOF) {n = 0; Len = strlen (s); For (INT I = len-1; s [I]! = '. '; I --) // processing before the decimal point {n/= double (8.0 ); // convert to N + = double (s [I]-'0');} n/= double (8.0 ); // The one before the decimal point has not processed printf ("% s", S); printf ("[8] = %. 45g [10] \ n ", n); // % G refers to the floating point number, remove meaningless zero} return 0 ;}